Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - General Problems - Page 47: 71

Answer

(a) The time to reach the bottom of the cliff is 5.80 seconds. (b) The speed just before hitting the bottom is 41.3 m/s. (c) The total distance the stone traveled is 99.6 meters.

Work Step by Step

(a) $y = y_0 + v_0t + \frac{1}{2}at^2$ $-75.0 = 0 + (15.5)t -4.90t^2$ $4.90 t^2 -15.5~t - 75.0 = 0$ We can use the quadratic formula to solve for t. $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $t = \frac{15.5 \pm \sqrt{(-15.5)^2 - (4)(4.90)(-75.0)}}{(2)(4.90)}$ $t = -2.64 s, 5.80 s$ The negative value for t is unphysical, so the time to reach the bottom of the cliff is 5.80 seconds. (b) $v = v_0 + at = 15.5 ~m/s - (9.80 ~m/s^2)(5.80 ~s)$ $v = -41.3 ~m/s$ The speed just before hitting the bottom is 41.3 m/s. (c) Let $y_{max}$ be the maximum height above the top of the cliff. $y_{max} = \frac{v^2-v_0^2}{2a} = \frac{0 - (15.5 ~m/s)^2}{(2)(-9.80 ~m/s^2)} = 12.3 ~m$ Then the total distance is 12.3 m + 12.3 m + 75.0 m = 99.6 meters.
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